For natural number n-n -2-nn- ifA- n-3 B- n - 3C- n - 4D- n-4

The correct option is B n≥ 3 Check through option, condition (n!)^2 gt; n^n is true when n ≥ 3. ...

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Standard XII

Mathematics

Proof by mathematical induction

For natural n...

Question

For natural number n, $(n!)^{2}$ > $n^{n}$ , if

n > 3

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n > 4

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$n \geq$ 4

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$n \geq$ 3

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Solution

The correct option is D
$n \geq$ 3

Check through option, condition $(n!)^{2}$ > $n^{n}$ is

true when n ≥ 3.

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Similar questions

Q. For natural number n,

(n!)^{2} > n^{n}

For natural number n, $(n!)^{2}$ > $n^{n}$ , if

Q. Statement-l: For every natural number

n \geq 2, \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \dots \dots + \frac{1}{\sqrt{n}} > \sqrt{n}

.
Statement-2: For every natural number

n \geq 2, \sqrt{n (n + 1)} < n + 1

Statement I

For every natural number

n \geq 2

\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{n}} > \sqrt{n}

Statement II

For every natural number

n \geq 2

\sqrt{n (n + 1)} < n + 1

Statement 1: For every natural number $n \geq 2.$
$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + . . . + \frac{1}{\sqrt{n}} > \sqrt{n} .$
Statement 2: For every natural number $n \geq 2,$
$\sqrt{n (n + 1)} < n + 1.$

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For natural number n, (n!)2 > nn, if

The correct option is D n≥ 3 Check through option, condition (n!)2 > nn is true when n ≥ 3.

For natural number n, $(n!)^{2}$ > $n^{n}$ , if

The correct option is D
$n \geq$ 3

Check through option, condition $(n!)^{2}$ > $n^{n}$ is

true when n ≥ 3.